Heisenberg Lie Algebras and Polynomial Weyl Algebras

By the matrix-product formula [F2] and trace definition [F1], $\mathrm{tr}(AB)={\sum }_{i<d}{\sum }_{j<d}{a}_{ij}{b}_{ji}$.

By [F1] and the matrix-product formula [F2], $\mathrm{tr}(BA)={\sum }_{i<d}{\sum }_{j<d}{b}_{ij}{a}_{ji}$. Since $R$ is commutative, reordering each product and exchanging the finite dummy indices gives $\mathrm{tr}(BA)={\sum }_{i<d}{\sum }_{j<d}{a}_{ij}{b}_{ji}=\mathrm{tr}(AB)$.

The diagonal entries of $AB-BA$ are the differences of the corresponding diagonal entries, so [F1] and step 2.1 give $\mathrm{tr}(AB-BA)=\mathrm{tr}(AB)-\mathrm{tr}(BA)=0$. The same calculation covers $d=0$ because both sums are empty.

Bilinearity is part of the bracket construction in [F1]. For $v={\sum }_i(a_i{X}_i+b_i{P}_i)+cZ$, its self-bracket is $[v,v]={\sum }_i(b_i{a}_i-a_i{b}_i)Z=0$, since multiplication in $k$ is commutative. Thus the bracket is alternating in every characteristic, including characteristic $2$.

By [F1], each bracket lies in $kZ$ and $Z$ brackets to zero with all of ${\mathfrak{h}}_n(k)$. Every nested bracket is therefore zero, so each term in the Jacobi sum vanishes. Together with step 1.1, the axioms in [F2] hold.

Choose a $k$-basis $e_1,\dots ,e_d$ of $M$ and let $D$ and $X$ be the matrices of the actions of $p_1$ and $x_1$. Writing $X{e}_j={\sum }_r{x}_{rj}{e}_r$ and $D{e}_r={\sum }_i{d}_{ir}{e}_i$, [F5] and the product formula [F6] show that the matrices of the composite actions $p_1{x}_1$ and $x_1{p}_1$ are $DX$ and $XD$, respectively. The scalar copy of $k$ acts through the given vector-space structure, so the module axiom [F2] and relation [F1] give $DX-XD=\mathrm{\hslash }{I}_d$, where $d={\dim }_{k}M\ge 1$.

By [F3], $\mathrm{tr}(DX-XD)=0$. By [F4] and [F6], $\mathrm{tr}(\mathrm{\hslash }{I}_d)=\mathrm{\hslash }d$, so $\mathrm{\hslash }d=0$ in $k$.

Characteristic zero makes the image of every positive integer $d$ nonzero in $k$. Since $\mathrm{\hslash }\mathrm{\ne }0$ and a field has no zero divisors, $\mathrm{\hslash }d\mathrm{\ne }0$, contradicting step 2.1. Hence no such nonzero module exists.

Order the basis of ${\mathfrak{h}}_n(k)$ as $X_1<\cdots <X_n<P_1<\cdots <P_n<Z$. By [F4], the elements $X^{\alpha }{P}^{\beta }{Z}^m$ form a $k$-basis of $U({\mathfrak{h}}_n(k))$. In particular, the powers $Z^m$ are linearly independent; let $k[Z]$ denote their polynomial subalgebra. Since $Z$ is central by [F1], grouping this basis by $\alpha ,\beta$ gives the direct-sum decomposition ${⨁}_{\alpha ,\beta \in {\mathbb{N}}^n}k[Z]X^{\alpha }{P}^{\beta }$ as a free $k[Z]$-module.

Send $X_i\mapsto x_i$, $P_i\mapsto p_i$, and $Z\mapsto \mathrm{\hslash }1$ in $W_n(k;\mathrm{\hslash })$. By [F1] and [F5], this linear assignment preserves the bracket on every basis pair, hence on all pairs by bilinearity. By [F9] it extends to a unital algebra map $T({\mathfrak{h}}_n(k))\to W_n(k;\mathrm{\hslash })$. The defining relators of $U({\mathfrak{h}}_n(k))$ in [F3] map to zero. The map's kernel is a two-sided ideal because it is an algebra homomorphism, so it contains their generated ideal by [F7], and [F10] factors the map through $U({\mathfrak{h}}_n(k))$. This induced algebra map sends $Z-\mathrm{\hslash }1$ to zero; its kernel is two-sided and contains the ideal $(Z-\mathrm{\hslash }1)$ generated by that central element by [F7], so a second application of [F10] gives $\varphi :Q_{\mathrm{\hslash }}\to W_n(k;\mathrm{\hslash })$. The factor maps remain $k$-linear because the quotient maps are surjective.

Orient the relations of [F5] by $x_j{x}_i\to x_i{x}_j$ and $p_j{p}_i\to p_i{p}_j$ when $j>i$, and $p_i{x}_j\to x_j{p}_i+\mathrm{\hslash }{\delta }_{ij}1$. For each word use the lexicographic measure (word length, number of inversions for $x_1<\cdots <x_n<p_1<\cdots <p_n$). Each swapped word has one fewer inversion, and each scalar correction has smaller length; hence every reduction terminates.

By step 1.1, $U({\mathfrak{h}}_n(k))$ is free as a $k[Z]$-module on $X^{\alpha }{P}^{\beta }$. Since $Z$ is central, the ideal $(Z-\mathrm{\hslash })U$ is the direct sum of $(Z-\mathrm{\hslash })k[Z]X^{\alpha }{P}^{\beta }$. For $f(Z)={\sum }_{m=0}^N{c}_m{Z}^m$, the identity $Z^m-{\mathrm{\hslash }}^m=(Z-\mathrm{\hslash }){\sum }_{r=0}^{m-1}{Z}^{m-1-r}{\mathrm{\hslash }}^r$ for $m\ge 1$ shows that $f(Z)-f(\mathrm{\hslash })$ is divisible by $Z-\mathrm{\hslash }$. The evaluation map ${\mathrm{ev}}_{\mathrm{\hslash }}:k[Z]\to k$, $f\mapsto f(\mathrm{\hslash })$ is surjective because it fixes constants, and the displayed identity shows its kernel is exactly $(Z-\mathrm{\hslash })$; hence $k[Z]\mathrm{/}(Z-\mathrm{\hslash })\cong k$. Therefore the classes of $X^{\alpha }{P}^{\beta }$ form a basis of $Q_{\mathrm{\hslash }}$.

The overlapping mixed reductions are $p_k{p}_j{x}_i$ with $k>j$ and $p_k{x}_j{x}_i$ with $j>i$. For $p_k{p}_j{x}_i$, reducing the left pair gives $p_k{p}_j{x}_i\to p_j{p}_k{x}_i\to p_j{x}_i{p}_k+\mathrm{\hslash }{\delta }_{ki}{p}_j\to x_i{p}_j{p}_k+\mathrm{\hslash }{\delta }_{ji}{p}_k+\mathrm{\hslash }{\delta }_{ki}{p}_j$, while reducing the right pair gives $p_k{p}_j{x}_i\to p_k{x}_i{p}_j+\mathrm{\hslash }{\delta }_{ji}{p}_k\to x_i{p}_k{p}_j+\mathrm{\hslash }{\delta }_{ki}{p}_j+\mathrm{\hslash }{\delta }_{ji}{p}_k\to x_i{p}_j{p}_k+\mathrm{\hslash }{\delta }_{ki}{p}_j+\mathrm{\hslash }{\delta }_{ji}{p}_k$; these final expressions agree. For $p_k{x}_j{x}_i$, reducing the left pair gives $p_k{x}_j{x}_i\to x_j{p}_k{x}_i+\mathrm{\hslash }{\delta }_{kj}{x}_i\to x_j{x}_i{p}_k+\mathrm{\hslash }{\delta }_{ki}{x}_j+\mathrm{\hslash }{\delta }_{kj}{x}_i\to x_i{x}_j{p}_k+\mathrm{\hslash }{\delta }_{ki}{x}_j+\mathrm{\hslash }{\delta }_{kj}{x}_i$, while reducing the right pair gives $p_k{x}_j{x}_i\to p_k{x}_i{x}_j\to x_i{p}_k{x}_j+\mathrm{\hslash }{\delta }_{ki}{x}_j\to x_i{x}_j{p}_k+\mathrm{\hslash }{\delta }_{kj}{x}_i+\mathrm{\hslash }{\delta }_{ki}{x}_j$; these final expressions also agree. For a strictly decreasing same-type triple, both routes $x_k{x}_j{x}_i\to x_j{x}_k{x}_i\to x_j{x}_i{x}_k\to x_i{x}_j{x}_k$ and $x_k{x}_j{x}_i\to x_k{x}_i{x}_j\to x_i{x}_k{x}_j\to x_i{x}_j{x}_k$ reach the same sorted word; the same calculation holds for $p$'s. Repeated indices remove the corresponding swap, and disjoint reductions commute by distributivity. Thus every local ambiguity joins.

By step 2.1, the central quotient $Q_{\mathrm{\hslash }}$ has the normal basis just described. In $Q_{\mathrm{\hslash }}$, the images of $X_i,P_i$ satisfy $[P_i,X_j]=\mathrm{\hslash }{\delta }_{ij}1$, and the same-type commutators vanish. Send the basis generators $x_i,p_i$ of $E_n$ to these images; [F9] extends this linear assignment to a unital algebra map $T(E_n)\to Q_{\mathrm{\hslash }}$. It kills the relators in [F5]. Its kernel is a two-sided ideal, so it contains their generated ideal by [F7], and [F10] factors it uniquely through the quotient to $\psi :W_n(k;\mathrm{\hslash })\to Q_{\mathrm{\hslash }}$. This factor map is $k$-linear because the quotient map is surjective.

Induct on the decreasing measure from step 1.3. If two reduction sequences first choose different redexes, step 2.2 joins those first reductions; every word in the joining expressions has smaller measure, so the induction hypothesis gives it a unique final reduction. Thus every word has one normal form. Because each defining relation is a permitted reduction in any left-right word context, confluence makes the normal-form map kill every such context and hence the whole ideal $J_{n,\mathrm{\hslash }}$ from [F5]. Each reduction also changes a word by an element of $J_{n,\mathrm{\hslash }}$, so every class is represented by its normal form. Therefore the quotient is spanned by normal words, and any linear relation among normal words in the quotient is fixed by the normal-form map while also lying in its kernel, so all its coefficients are zero.

The composites $\varphi \psi$ and $\psi \varphi$ fix every generator and the unit of their respective algebras. Those generators generate each quotient, so the composites are identity maps and $\varphi ,\psi$ are inverse isomorphisms. The basis from step 3.2 therefore transfers to $W_n(k;\mathrm{\hslash })$.

If $\mathrm{\hslash }\mathrm{\ne }0$, the assignments $x_i\mapsto x_i$, $p_i\mapsto \mathrm{\hslash }{p}_i$ preserve the relations from $W_n(k;\mathrm{\hslash })$ to $A_n(k)$; the inverse rescales $p_i$ by ${\mathrm{\hslash }}^{-1}$. If $\mathrm{\hslash }=0$, [F5] makes every pair of generators commute. Thus the parameter branches and normal basis stated in the theorem hold.

Fix $a,b$ and $\xi \in D$, abbreviate $U=U_a$ and $c=c_{ab}$, and set $F(s)=U(s)A_{b}U(-s)\xi$. Every product is defined: $U(-s)\xi \in D$, $A_{b}D\subset D\subset D(A_a)$, and $A_{a}D\subset D\subset D(A_b)$. The graph-orbit hypothesis gives the norm derivative of $\eta (s)=A_{b}U(-s)\xi$ as ${\eta }^{\mathrm{\prime }}(s)=-i{A}_b{A}_{a}U(-s)\xi$. The difference quotient for $U(s)\eta (s)$ splits into $U(s+h)(\eta (s+h)-\eta (s))\mathrm{/}h$ and $(U(s+h)-U(s))\eta (s)\mathrm{/}h$. Strong continuity and norm one of $U(s+h)$ handle the first term; [F1] applies to the fixed vector $\eta (s)\in D(A_a)$ in the second. Consequently $F^{\mathrm{\prime }}(s)=iU(s)(A_a{A}_b-A_b{A}_a)U(-s)\xi =-c\xi$. The curve $G(s)=F(s)+cs\xi$ is norm differentiable on $\mathbb{R}$ with derivative zero, hence is continuous there; its real-variable norm derivative is also its Fréchet derivative, since the remainder divided by $\mathrm{\mid }h\mathrm{\mid }$ tends to zero. For any $s$, apply [F5] to $G$ on the closed interval with endpoints $0,s$ (the case $s=0$ is immediate). It gives $G(s)=G(0)=A_b\xi$, so $F(s)=A_b\xi -cs\xi$.

For fixed $s$, the closed operator $B=U(s)A_{b}U(-s)$ has domain $U(s)D(A_b)$ and is self-adjoint: substituting $U(-s)x,U(-s)y$ in the adjoint-domain condition gives $B^{\ast }=U(s)A_b^{\ast }U(-s)=B$. Moreover $D$ is a core for $B$, because $U(s)$ transports graph approximations from the core $D$ of $A_b$ and $U(s)D=D$. The operator $A_b-csI$ is closed on $D(A_b)$: convergence of $x_j$ and $(A_b-csI)x_j$ implies convergence of $A_b{x}_j$ and closedness of $A_b$ applies. The two graph norms of $A_b$ and $A_b-csI$ are equivalent by the triangle inequality, so $D$ is a core for this shifted operator as well. Step 1.1 identifies their restrictions to $D$; taking their graph closures yields the exact operator equality $B=A_b-csI$, including equality of domains. In particular $U(s)D(A_b)=D(A_b)$ has now been proved, rather than assumed.

Fix $s$ and consider $K(t)=U_a(s)U_b(t)U_a(-s)$ and $L(t)=e^{-i{c}_{ab}st}{U}_b(t)$. Both are strongly continuous unitary groups by their formulas and [F1]. For $K$, taking difference quotients after unitary conjugation shows its exact derivative domain is $U_a(s)D(A_b)$, with derivative $iB$. For $L$, differentiating the scalar factor shows that every vector of $D(A_b)$ has derivative $i(A_b-c_{ab}sI)$. Conversely, if the derivative of $L(t)x$ at zero exists, writing $U_b(t)x=e^{i{c}_{ab}st}L(t)x$ proves the derivative of $U_b(t)x$ exists, so $x\in D(A_b)$ by [F1]. Thus the exact generators agree by step 2.1; [F2] gives $K=L$. Multiplying on the right by $U_a(s)$ proves the claimed pairwise Weyl relation on all of $H$. This is the functional-calculus exponentiation step; no power series of unbounded operators was used.

All factors defining $W(f)$ are unitary, and $W(0)=I$. Their finite product is strongly continuous jointly in $f$: subtract the products at two points and telescope, using norm one for the factors and strong continuity on each fixed vector in the finitely many resulting terms. To multiply $W(f)$ and $W(g)$, move each $U_a(g_a)$ left past $U_b(f_b)$ for $b>a$. Step 3.1 contributes $\exp (i{c}_{ab}{f}_b{g}_a)$ for each such pair; same-index factors combine by the group law. The scalar exponent after dividing by the defining scalar of $W(f+g)$ is $\frac{i}{2}{\sum }_{a<b}{c}_{ab}(f_a{f}_b+g_a{g}_b-(f_a+g_a)(f_b+g_b))+i{\sum }_{a<b}{c}_{ab}{f}_b{g}_a=-\frac{i}{2}{\sum }_{a<b}{c}_{ab}(f_a{g}_b-f_b{g}_a)=-i\sigma (f,g)\mathrm{/}2$. This proves the full displayed Weyl relation. Since $\sigma (f,-f)=0$, it gives $W(f)W(-f)=W(-f)W(f)=I$, so unitarity yields $W(f{)}^{\ast }=W(-f)$.

In the stated block-matrix case the commutators on $D$ are $[Q_j,Q_k]=[P_j,P_k]=0$ and $[Q_j,P_k]=i\kappa {\delta }_{jk}I$. Hence $[(i\mathrm{/}\kappa )P_j,Q_k]={\delta }_{jk}I$ when $\kappa \mathrm{\ne }0$. Domain invariance makes these restrictions endomorphisms of $D$. Since $D$ is a complex vector space, pointwise addition and scalar multiplication and composition make ${\mathrm{End}}_{\mathbb{C}}(D)$ a unital associative $\mathbb{C}$-algebra: composition is associative, the identity map is its unit, and $\mathbb{C}$-bilinearity follows by evaluating on vectors. Regard $E_n$ as the span of the generators $x_j,p_j$ and send them to $Q_j{\mathrm{\mid }}_D$ and $(i\mathrm{/}\kappa )P_j{\mathrm{\mid }}_D$. By [F6] this extends to a unital algebra map from $T(E_n)$. The displayed commutators show that this map kills each defining relator in [F4]; its kernel is a two-sided ideal because it is an algebra homomorphism, so it contains the generated ideal by [F9]. Then [F7] factors it through $A_n(\mathbb{C})$; write $\rho$ for this factor map. The factor is $\mathbb{C}$-linear because the quotient map is surjective, and evaluation $a\cdot \xi :=\rho (a)\xi$ on $D$ gives a unital left module by [F8]. Step 4.1 gives its eligible exponentiated realization, with the exact phase and finite-mode scope claimed. The proof uses Choice through [F1]–[F2]'s spectral/Stone construction and Countable Choice in the self-adjoint/graph supplier closure [F3]; it makes no assertion that the common-core and graph-orbit assumptions follow from algebraic relations alone.